Products of Digits

This problem was recently featured as a Problem of the Week. My solution (slightly modified) is below.

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For any single-digit number, the number is equal to the product of its digits.

Now, consider the $(n+1)$ -digit integer $N=\overline{a_n a_{n-1}\cdots a_1 a_0}$ (where $n>0$ and $a_n\neq 0$ ). Note that $0\leq a_i \leq 9$ for all $i$ . $\begin{aligned} N &= \sum_{i=0}^n 10^i a_i \\ &\geq 10^n a_n \\ &> a_n\prod_{i=0}^{n-1} a_i \\ &= \prod_{i=0}^{n} a_i \end{aligned}$

Thus, if $N\geq 10$ , the product of the digits will always be strictly less than $N$ . In all cases, the product of the digits is at least the number itself, and the answer is True .

$\overline{x_n \dots x_1} \geq x_n 10^{n-1} > x_n 9^{n-1} \geq x_n.x_{n-1}\dots x_1$

$A||B||C||......\geq A\times B\times C......$ where $A, B, C$ are single digit positive integers.

$(||)$ denotes concatenation.

Its equal to for all positive single digit integers and greater for all more 2+ digits integers.

Hence, $\boxed{TRUE}.$